# Eigenspaces and jordan normal form

I have a question here regarding the jordan normal form of two matrices where the eigenspace is one is contained in the other.

Let $A,B$ be two $n \times n$ matrices s.t $AB=BA$. I firstly proved that the eigenspace of A is preserved by B, getting the result that $BV(\lambda) \subset V(\lambda)$ where $V(\lambda) = \{ x | Ax = \lambda x\}$.

I assume $A$ is diagonalisable and have to prove that there exists an invertible matrix $T$ s.t both $TBT^{-1}$ and $TAT^{-1}$ are in jordan normal form. Since $A$ is diagonalisable, it is similar to a diagonal matrix. So this means that there exists invertible $T$ such that $TAT^{-1}$ is diagonal, which I believe is also its Jordan normal form just with jordan blocks of size $1$.

Since the eigenspaces of $B$ is contained in $A$, then an eigenvalue of $B$ is necessarily an eigenvalue of $A$, so it will have the same eigenvalues, but not necessarily as many jordan blocks. How can i further go onto prove that $TBT^{-1}$ is in Jordan normal form?

## 1 Answer

This won't quite work: e.g. you might have $A = I$, in which case your $T$ could be any invertible matrix, and there's no reason to think $T B T^{-1}$ will be in Jordan normal form. What you want to do is choose $T$ on each eigenspace of $A$ to put into Jordan form the restriction of $B$ to that eigenspace.

• Can i instead use the fact that $TAT^{-1}=D$ (where D is a diagonal matrix, T some invertible matrix) along with $AB = BA$ to get a similar expression for B? The only problem is i still cannot seem to sufficiently show its in JNF. – Vinny Jan 27 '15 at 13:58
• That's the point. Once you have $A$ diagonalized, you have its eigenspaces (corresponding to blocks of indices where $D$ has each eigenvalue of $A$ as diagonal element); $TBT^{-1}$ consists of blocks on the diagonal corresponding to these, and you put each of these into Jordan form. – Robert Israel Jan 27 '15 at 15:44